What is the time complexity of Minimum Cost Walk in Weighted Graph?

The optimal solution has a time complexity of O(Q * E * A * log*N).

What is the space complexity of Minimum Cost Walk in Weighted Graph?

The space complexity is O(N + E + Q).

What data structures are used to solve Minimum Cost Walk in Weighted Graph?

This problem uses Array, Bit Manipulation, Union Find, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Cost Walk in Weighted Graph?

Insight 1: If two vertices are not connected, the minimum cost is -1. We can efficiently determine connectivity using the Union-Find (DSU) data structure. Insight 2: The bitwise AND operation implies that the cost can only decrease (or stay the same) as we add more edges to the walk. Therefore, we can iterate on potential costs which are bitmasks present in the edge weights.

Which companies ask LeetCode 3108. Minimum Cost Walk in Weighted Graph in interviews?

This problem is frequently asked at DE Shaw.

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Minimum Cost Walk in Weighted Graph - LeetCode 3108 Solution

Minimum Cost Walk in Weighted Graph - Complete Solution Guide

Minimum Cost Walk in Weighted Graph is LeetCode problem 3108, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is an undirected weighted graph with n vertices labeled from 0 to n - 1 . You are given the integer n and an array edges , where edges[i] = [u i , v i , w i ] indicates that there is an edge between vertices u i and v i with a weight of w i . A walk on a graph is a sequence of vertices and edges. The walk starts and ends with a vertex, and each edge connects the vertex that comes before it and the vertex that comes after it. It's important to note that a walk may visit the same edge or ver

Detailed Explanation

The problem asks us to find the minimum cost of a walk between two given vertices in a weighted undirected graph for multiple queries. The cost of a walk is defined as the bitwise AND of the weights of the edges traversed during the walk. If no walk exists between the given vertices, the cost is -1. The graph has 'n' vertices labeled from 0 to n-1, and we are provided with edges in the form of [u, v, w], where 'u' and 'v' are the vertices connected by the edge, and 'w' is the weight of the edge. We are also given multiple queries in the form of [s, t], where 's' is the starting vertex and 't' is the ending vertex.

Solution Approach

The solution utilizes an iterative approach with Union-Find to determine the minimum cost of a walk between two vertices for each query. It starts by checking if a path exists between the source and target vertices using a DSU. If a path exists, the minimum cost is initialized to 0. The solution then iteratively refines the minimum cost by considering different bitmasks. In each iteration, edges that are 'supermasks' of the current cost are used to build connected components using DSU. The minimum weight within each component is calculated, and the minimum cost for each query is updated. This process continues until no further updates are possible, or, effectively, the minimum possible cost is found.

Step-by-Step Algorithm

Step 1: Initialize the answer array with -1 for all queries.
Step 2: Use a DSU (Union-Find) to check if a path exists between the source and target vertices for each query. If a path exists, initialize the corresponding answer to 0, otherwise keep it as -1.
Step 3: Create a list of 'active queries' for the queries which has a path (initially identified as '0' min cost).
Step 4: Iteratively improve the answer for the active queries. This involves grouping active queries by their current minimum costs.
Step 5: For each group, build a DSU and include only the edges whose weights are 'supermasks' of the group's 'cost_mask' .
Step 6: Calculate the minimum weight within each connected component (using the supermask edges)
Step 7: Update the answers for queries in this group with the component cost. If any answer changes, then continue with active queries which got updated.
Step 8: Repeat from step 4 until there are no changes to update the active queries or until all queries are correctly processed.
Step 9: Return the answer array.

Key Insights

Insight 1: If two vertices are not connected, the minimum cost is -1. We can efficiently determine connectivity using the Union-Find (DSU) data structure.
Insight 2: The bitwise AND operation implies that the cost can only decrease (or stay the same) as we add more edges to the walk. Therefore, we can iterate on potential costs which are bitmasks present in the edge weights.
Insight 3: The key idea behind the iterative improvement is that we can consider edges whose weight 'w' is a supermask of the current cost bitmask 'cost_mask'. We can then determine connected components using only those edges, and find the minimum weight within those components.
Insight 4: The number of iterations required is bounded by the number of bits in the largest possible weight (10^5). Since the maximum weight is 10^5, we will have at most ~17 iterations.

Complexity Analysis

Time Complexity: O(Q * E * A * log*N)

Space Complexity: O(N + E + Q)

Topics

This problem involves: Array, Bit Manipulation, Union Find, Graph.

Companies

Asked at: DE Shaw.

Minimum Cost Walk in Weighted Graph | LeetCode 3108

Hard

Array Bit Manipulation Union Find Graph

Time: O(Q * E * A * log*N)Space: O(N + E + Q)

Solution Code

Available in:

import java.util.*;

class DSU {
    int[] parent;

    public DSU(int n) {
        parent = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
    }

    public int find(int i) {
        if (parent[i] == i) {
            return i;
        }
        parent[i] = find(parent[i]);
        return parent[i];
    }

    public boolean union(int i, int j) {
        int root_i = find(i);
        int root_j = find(j);
        if (root_i != root_j) {
            parent[root_j] = root_i;
            return true;
        }
        return false;
    }
}

class Solution {
    public int[] minimumCost(int n, int[][] edges, int[][] query) {
        int q_len = query.length;
        int[] ans = new int[q_len];
        Arrays.fill(ans, -1);
        
        // Identify active queries (where s and t are connected in the original graph)
        DSU initial_dsu = new DSU(n);
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int _w = edge[2];
            initial_dsu.union(u, v);
        }

        List<Integer> active_queries = new ArrayList<>();
        for (int i = 0; i < q_len; i++) {
            int s = query[i][0];
            int t = query[i][1];
            if (initial_dsu.find(s) == initial_dsu.find(t)) {
                ans[i] = 0;
                active_queries.add(i);
            }
        }

        while (true) {
            boolean updated_in_iteration = false;
            Map<Integer, List<Integer>> groups = new HashMap<>();
            for (int q_idx : active_queries) {
                if (!groups.containsKey(ans[q_idx])) {
                    groups.put(ans[q_idx], new ArrayList<>());
                }
                groups.get(ans[q_idx]).add(q_idx);
            }
            
            List<Integer> next_active_queries = new ArrayList<>();

            for (Map.Entry<Integer, List<Integer>> entry : groups.entrySet()) {
                int cost_mask = entry.getKey();
                List<Integer> q_indices = entry.getValue();
                DSU dsu = new DSU(n);
                
                // Build graph with edges that are supermasks of cost_mask
                List<int[]> relevant_edges = new ArrayList<>();
                for (int[] edge : edges) {
                    int u = edge[0];
                    int v = edge[1];
                    int w = edge[2];
                    if ((w & cost_mask) == cost_mask) {
                        dsu.union(u, v);
                        relevant_edges.add(edge);
                    }
                }
                
                Map<Integer, Integer> component_costs = new HashMap<>();
                for(int i = 0; i < n; i++){
                    component_costs.put(i, (1 << 30) - 1);
                }

                for (int[] edge : relevant_edges) {
                    int u = edge[0];
                    int v = edge[1];
                    int w = edge[2];
                    int root = dsu.find(u);
                    component_costs.put(root, component_costs.get(root) & w);
                }

                boolean new_active_in_group = false;
                for (int q_idx : q_indices) {
                    int s = query[q_idx][0];
                    int root_s = dsu.find(s);
                    int new_cost = component_costs.get(root_s);
                    
                    if (ans[q_idx] != new_cost) {
                        ans[q_idx] = new_cost;
                        updated_in_iteration = true;
                        new_active_in_group = true;
                    }
                }

                if (new_active_in_group) {
                    next_active_queries.addAll(q_indices);
                }
            }

            if (!updated_in_iteration) {
                break;
            }
            
            active_queries = next_active_queries;
        }
        
        return ans;
    }
}

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Problem Info

Description

Problem 3108: Minimum Cost Walk in Weighted Graph - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Union Find Graph

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3108

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize the answer array with -1 for all queries.
2
Step 2: Use a DSU (Union-Find) to check if a path exists between the source and target vertices for each query. If a path exists, initialize the corresponding answer to 0, otherwise keep it as -1.
3
Step 3: Create a list of 'active queries' for the queries which has a path (initially identified as '0' min cost).
4
Step 4: Iteratively improve the answer for the active queries. This involves grouping active queries by their current minimum costs.
5
Step 5: For each group, build a DSU and include only the edges whose weights are 'supermasks' of the group's 'cost_mask' .
6
Step 6: Calculate the minimum weight within each connected component (using the supermask edges)
7
Step 7: Update the answers for queries in this group with the component cost. If any answer changes, then continue with active queries which got updated.
8
Step 8: Repeat from step 4 until there are no changes to update the active queries or until all queries are correctly processed.
9
Step 9: Return the answer array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(Q * E * A * log*N)

💾

Space Complexity

O(N + E + Q)

📝

Why?

The initial DSU operation takes O(E * alpha(N)) time. The main loop iterates at most O(A) times. Inside the loop, building the DSU takes O(E * alpha(N)) time in each iteration. Finding the components for each query takes O(Q). Updating ans array takes O(Q). Therefore, inside each iteration, the time complexity becomes O((Q+E) * alpha(N)). Overall TC is O(Q * E * A * alpha(N)). The space complexity is O(N) for DSU parent array, O(E) for the relevant edge list and O(Q) for storing active queries and groups.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dijkstra's Algorithm with Bitwise AND

Instead of the iterative approach, we can try to adapt Dijkstra's algorithm to this problem. We can maintain a priority queue of (cost, node) pairs, where cost is the minimum cost encountered so far to reach the given node. When relaxing an edge, we perform a bitwise AND of the current cost with the edge weight.

Time: O(E * log(V) * log(MaxWeight)), where MaxWeight is the maximum edge weight.Space: O(V + E)

Trade-offs: This approach may be simpler to implement, but the complexity is likely higher due to the log factor introduced by Dijkstra's. This might be useful if the graph is very sparse and the edge weights are large, where Dijkstra's can perform better.

✨

Key Insights

1Insight 1: If two vertices are not connected, the minimum cost is -1. We can efficiently determine connectivity using the Union-Find (DSU) data structure.
2Insight 2: The bitwise AND operation implies that the cost can only decrease (or stay the same) as we add more edges to the walk. Therefore, we can iterate on potential costs which are bitmasks present in the edge weights.
3Insight 3: The key idea behind the iterative improvement is that we can consider edges whose weight 'w' is a supermask of the current cost bitmask 'cost_mask'. We can then determine connected components using only those edges, and find the minimum weight within those components.
4Insight 4: The number of iterations required is bounded by the number of bits in the largest possible weight (10^5). Since the maximum weight is 10^5, we will have at most ~17 iterations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: The graph is not connected. The initial DSU check addresses this by setting the answer to -1 if there's no path between source and target.
Edge case 2: The cost of a walk can be 0. The algorithm handles this by initializing the ans array and actively updating it during each iteration.
Edge case 3: There are multiple edges between two vertices with different weights. The algorithm considers all edges during the DSU construction.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking for the initial connectivity between the source and target vertices, leading to incorrect results.

✗Common mistake 2: Incorrectly calculating the bitwise AND of edge weights, causing incorrect minimum cost calculations.

✗Common mistake 3: Not handling the case where no walk exists between two vertices, leading to incorrect results.

✗Common mistake 4: Forgetting to perform the bitwise AND to get the component costs. Always remember that the walk is defined by the AND operation.

✗Common mistake 5: Getting confused with the condition that an edge `(u,v,w)` is relevant based on `(w & cost_mask) == cost_mask`. Many beginners forget the AND operator precedence.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing where the cost represents available bandwidth or security level, and the minimum cost path needs to be determined while considering the bitwise AND of the bandwidth/security levels along the path.

🏦Real-world scenario 2: Supply chain optimization where the cost represents certain qualities of materials or resources (e.g., purity, strength), and the minimum cost path ensures a certain minimum level of all qualities are met.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Cost Walk in Weighted Graph"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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